MS_Kekas wrote: Positive integers $x, y$ satisfy the following conditions: $$\{\sqrt{x^2 + 2y}\}> \frac{2}{3}; \hspace{10mm} \{\sqrt{y^2 + 2x}\}> \frac{2}{3}$$ Show that $x = y$. Here $\{x\}$ denotes the fractional part of $x$. For example, $\{3.14\} = 0.14$. Proposed by Anton Trygub

Otherwise, w.l.o.g. $x<y$. Then $\sqrt{x^2+2y}>x+1$ and hence the conditions tell us that $\sqrt{x^2+2y}>x+\frac{5}{3}$ and $\sqrt{y^2+2x}>y+\frac{2}{3}$. But this implies $y>\frac{5}{3}x$ and $x>\frac{2}{3}y$ which is a contradiction as $\frac{5}{3} \cdot \frac{2}{3}>1$.